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R 2 D ⊆ R 2 f : D → R (x0, y0) ∈ D
f(x0, y0) = 0 (x0, y0) ∂yf(x, y) ∂yf(x, y) = 0
U V R x0 y0 ϕ : U → V
{(x, y) ∈ D : f(x, y) = 0} ∩ (U × V ) = {(x, ϕ(x)) : x ∈ U}, ϕ(x0) = y0.
ϕ f x x0
f (x0, y0)
ϕ x0
ϕ ′ (x0) = − ∂xf(x0, y0)
ϕyf(x0, y0) .
f (x0, y0)
ϕ ′ (x) = − ∂xf(x, ϕ(x))
, ∀x ∈ U.
∂yf(x, ϕ(x))
X, Y, Z D X × Y f : D → Z
(x0, y0) ∈ D f(x0, y0) = 0 (x0, y0)
∂Y f(x, y) ∂Y f(x0, y0) Y Z U ⊂ X V ⊂ Y
x0 y0 ϕ : U → V
{(x, y) ∈ D : f(x, y) = 0} ∩ (U × V ) = {(x, ϕ(x)) : x ∈ U}, ϕ(x0) = y0.
ϕ f x x0 f
(x0, y0)
ϕ ′ −1 (x0) = − ∂Y f(x0, y0) ◦ ∂Xf(x0, y0).
y = ϕ(x)
x = 2
x 2 + y 3 − 2xy − 1 = 0,
ϕ(2) = 1
f(x, y) = x 2 + y 3 − 2xy − 1 P = (2, 1) f(P ) = 0
∂yf(x, y) = 3y 2 − 2x ∂yf(P ) = −1 = 0
ϕ 2 F (x, ϕ(x)) = 0 ϕ(2) = 1 f ∈ C 1
ϕ C 1
ϕ ′ (2) = 2
d 2
d
dx ϕ(x) = −∂xf(x, ϕ(x)) 2x − 2ϕ(x)
= −
∂yf(x, ϕ(x)) 3ϕ2 (x) − 2x ,
dx2 ϕ(x) = −(2 − 2ϕ′ (x))(3ϕ2 (x) − 2x) − (2x − 2ϕ(x))(6ϕ(x)ϕ ′ (x) − 2)
(3ϕ2 (x) − 2x) 2
x = 2 ϕ(2) = 1 ϕ ′ (2) = 2 ϕ ′′ (2) = 1/16
,